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Defining on $R^3$, $V = \iiint_S dx \, dy \, dz $ as the volume of surface $S$, with $S$ closed, bounded and arc-connected. Which is the $S$ of minimal area, that contains $V$. I know it's a bit general, so maybe you could think of another restriction without losing too much generality

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In equation $V = \iiint_S dx dy dz$ it seems that $V$ is a number. So how can $S$ "contain" $V$? How can $V$ be a "long and narrow circular cylinder"? –  GEdgar Oct 6 '12 at 23:53
I think John is looking for the surface of minimal area whose volume is V, for any given V. –  Bananas Oct 7 '12 at 0:00
@GEdgar: I misinterpreted the question. –  Fernando Martin Oct 7 '12 at 0:09
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2 Answers

up vote 3 down vote accepted

The link given by Brad is the isoperimetric inequality, which states that of all surfaces with area $A$, the one enclosing the most volume is a sphere. This also implies the answer to your question: of all surfaces enclosing a given volume $V$, the one with least surface area is a sphere.

To see this, fix a volume $V$ and let $S$ be any surface enclosing volume $V$. Let $S_0$ be a sphere having the same surface area as $S$. By the isoperimetric inequality, $S_0$ encloses more volume than $S$ (or the same amount). So if we scaled down $S_0$ to get a smaller sphere $S_1$ whose volume is $V$, it would have smaller surface area than $S_0$, and hence smaller surface area than $S$.

So we have shown that the sphere $S_1$ whose volume is $V$ has smaller surface area than any other surface $S$ enclosing volume $V$.

The relationship between statements like this is called duality.

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The minimal surface is a sphere.

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I'm sorry I don't undertand that advanced math, but I think it says on Isoperimetric inequality in higher dimensions, given a surface area the max volume is a sphere, I'm asking given a volume the min area. –  John Oct 6 '12 at 22:53
@John: See my answer. –  Nate Eldredge Oct 7 '12 at 0:53
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